MIT OpenCourseWare http ocw mit edu 18 02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use visit http ocw mit edu terms 18 02 Lecture 14 Thu Oct 11 2007 Handouts PS5 solutions PS6 practice exams 2A and 2B Non independent variables Often we have to deal with non independent variables e g f P V T where P V nRT Question if g x y z c then can think of z z x y What are z x z y Example x2 yz z 3 8 at 2 3 1 Take di erential 2x dx z dy y 3z 2 dz 0 i e 4 dx dy 6 dz 0 constraint g c or dz 46 dx 16 dy So z x 4 6 2 3 and z y 1 6 taking the coe cients of dx and dy Or equivalently if y is held constant then we substitute dy 0 to get dz 4 6 dx so z x 4 6 2 3 In general g x y z c gx dx gy dy gz dz 0 If y held xed get gx dx gz dz 0 i e dz gx gz dx and z x gx gz Warning notation can be dangerous For example f x y x y f x 1 Change of variables x u y u v then f 2u v f u 2 x u but f x f u This is because f x means change x keeping y xed while f u means change u keeping v xed i e change x keeping y x xed f When there s ambiguity we must precise what is held xed deriv x with y held x y f xed deriv u with v held xed u v f f f We now have u v x v x y In above example we computed z x y When there is no risk of confusion we keep the old notation by default x means we keep y xed Example area of a triangle with 2 sides a and b making an angle is A 12 ab sin Suppose it s a right triangle with b the hypothenuse then constraint a b cos 3 ways in which rate of change of A w r t makes sense A 1 view A A a b independent variables usual A with a and b held xed This answers the question a and b xed changes triangle stops being a right triangle what happens to A 2 constraint a b cos keep a xed change while b does what it must to satisfy the A constraint a 3 constraint a b cos keep b xed change while a does what it must to satisfy the A constraint b How to compute e g A a treat A as function of a and while b b a 1 2 2 0 Substitution a b cos so b a sec A 12 ab sin 12 a2 tan A a 2 a sec Easiest here but it s not always possible to solve for b 1 Total di erentials da 0 a xed dA A d Aa da Ab db 12 ab cos d 12 b sin da and constraint da cos db b sin d Plugging in da 0 we get db b tan d 1 2 a sin db 1 2 and then 1 1 dA ab cos a sin b tan d 2 2 A a 1 1 1 ab cos a sin b tan ab sec 2 2 2 2 Chain rule A a A a Aa a a Ab b b A Ab b a We nd b a by using the constraint equation Ran out of time here Implicit di erentiation of constraint a b cos we have 0 a a b a cos b sin so b a b tan and hence A 1 1 1 ab cos a sin b tan ab sec a 2 2 2 The two systematic methods essentially involve calculating the same quantities even though things are written di erently 18 02 Lecture 15 Fri Oct 12 2007 Review topics Functions of several variables contour plots Partial derivatives gradient approximation formulas tangent planes directional derivatives Note partial di erential equations equations involving partial derivatives of an unknown function are very important in physics E g heat equation f t k 2 f x2 2 f y 2 2 f z 2 describes evolution of temperature over time Min max problems critical points 2nd derivative test checking boundary least squares won t be on the exam Di erentials chain rule change of variables Non independent variables Lagrange multipliers and constrained partial derivatives Re explanation of how to compute constrained partials say f f x y z where g x y z c To nd f z y 1 using di erentials df fx dx fy dy fz dz We set dy 0 since y held constant and want to eliminate dx For this we use the constraint dg gx dx gy dy gz dz 0 so setting dy 0 we get dx gz gx dz Plug into df df fx gz gx dz gz dz so f z y fx gz gx gz f f x f y f z x 2 using chain rule fx fz while z y x z y y z y z z y z y g x g y g z x g 0 gx gz z y x z y y z y z z y z y which gives x z y and hence the answer
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